OP Malhotra Class 11 Maths Solutions Chapter 23 Parabola Ex 23

Effective S Chand ISC Maths Class 11 Solutions Chapter 23 Parabola Ex 23 can help bridge the gap between theory and application.

S Chand Class 11 ICSE Maths Solutions Chapter 23 Parabola Ex 23

Question 1.
The focus at (10, 0) the directrix x = -10.
Solution:
Given focus is at (10, 0)
∴ axis of parabola along x-axis and its eqn. can be taken as y² = 4ax [Here a = 10]
∴ y² = 40x be the required eqn. of parabola.

Question 2.
The focus at (0, 5), the directrix y = – 5.
Solution:
Given focus is at (0, 5), on comparing with (0, a) ∴ a = 5

Thus axis of parabola is along y-axis.
Hence required eqn. of parabola be x² = (4 × 5)y i.e. x² = 20y

Question 3.
The focus at (- 3,0), the directrix x + 5 = 0.
Solution:
Let P (x, y) be any point on the parabola.
Then by definition, we have
| PF | = | PM |
\(\sqrt{(x-2)^2+(y+3)^2}\) = \(\frac{|x+5|}{1}\)
On squaring both sides ; we have
(x + 3)² + y² = (x + 5)²
⇒ x² + 6x + 9 + y² = x² + 10x + 25
⇒ y² = 4x + 16 = 4 (x + 4)

Question 4.
The focus at (2, – 3), the directrix x + 5 = 0.
Solution:
Given focus be F (2, – 3) and eqn. of directrix be x + 5 = 0
Let P (x, y) be any point on parabola.
\(\sqrt{(x-2)^2+(y+3)^2}\) = \(\frac{|x+5|}{1}\)
On squaring both sides ; we have
(x – 2)² + (y + 3)² = (x + 5)²
⇒ x² – 4x + 4 + y² + 6y + 9 = x² + 10x + 25
⇒ y² – 14x + 6y – 12 = 0,
which is the required eqn. of parabola.

Question 5.
The focus at (1, 1), the directrix x – y = 3.
Solution:
Given focus be F (1, 1) and eqn. of given
directrix be x-y -3 = 0
Let P (x, y) be any point on parabola
then by def. | PF | = | PM |
\(\sqrt{(x-1)^2+(y-1)^2}\) = \(\frac{|x-y-3|}{\sqrt{1^2+(-1)^2}}\)
On squaring both sides ; we have
(x – 1)² + (y – 1)² = \(\frac{|x-y-3|^2}{2}\)
⇒ 2 [(x – 1)² + (y – 1)²] = (x – y – 3)²
⇒ 2x² + 2y² – 4x – 4y + 4 = x² – y² + 9 – 2xy + 6y – 6x
⇒ x² + y² + 2xy + 2x – 10y – 5 = 0
which is the required eqn. of parabola.

Question 6.
The vertex at the origin, the axis along the x-axis, and passes through (- 3, 6).
Solution:
Let the eqn. of parabola whose vertex is at (0,0) and axis along x-axis be given by
y² = 4ax …(1)
eqn. (1) passes through the point (-3,6)
∴ 36 = 4a (- 3) ⇒ a = – 3
Thus eqn. (1) reduces to ; y² = – 12x be the required eqn. of parabola.

Question 7.
The focus at (- 2, -1) and the latus rectum joins the points (- 2, 2) and (- 2, – 4).
Solution:
Given focus of required parabola be (- 2, -1) and end points of latus rectum are (- 2, 2) and (- 2, – 4)
since x-coordinates of both end points be same.
∴ axis of the parabola be parallel to x-axis Let the eqn. of parabola be taken as :
(y – α)² = ± 4a (x – β) …(1)
length of latus rectum = 4a
= \(\sqrt{(-2+2)^2+(-4-2)^2}\) = 6
Thus eqn. (1) reduces to ;
(y – α)² = ± 6 (x – β) …(2)
The points (- 2, 2) and (- 2, – 4) lies on eqn. (2); we get
(2 – α)² = ± 6 (- 2 – β) …(3)
(- 4 – α)² = ± 6 (- 2 – β) …(4)

On dividing (3) by (4); we have
(2 – α)² = (a + 4)²
⇒ α² – 4α + 4 = α² + 8α + 16
⇒ α = – 1
∴ from (3); 9 = ± 6 (- 2 – α)
⇒ \(\frac { 3 }{ 2 }\) = ± (- 2 – β)
Case-I. \(\frac { 3 }{ 2 }\) = (- 2 – β)
⇒ β = – 2 – \(\frac { 3 }{ 2 }\) = –\(\frac { 7 }{ 2 }\)

Case-II. \(\frac { 3 }{ 2 }\) = – (- 2 – β)
⇒ \(\frac { 3 }{ 2 }\) = 2 + β ⇒ β = – \(\frac { 1 }{ 2 }\)
∴ from (1); we have
(y + 1)² = + 6\(\left(x+\frac{7}{2}\right)\)
⇒ y² + 2y – 6x – 20 = 0
and (y + 1)² = – 6\(\left(x+\frac{1}{2}\right)\)
⇒ y² + 2y – 6x + 4 = 0

Question 8.
The vertex at (- 2,3) and the focus at (1,3).
Solution:
Clearly the axis of the parabola be parallel to x-axis and its eqn. can be taken as,
(y – β)² = 4a (x – α) …(1)

given vertex of parabola be (- 2, 3).
∴ eqn. (1) reduces to
(y – 3)² = 4a (x + 2) …(2)
a – distance between focus and vertex
= 1 – (- 2) = 3
∴ eqn. (2) reduces to ;
(y – 3)² = 12 (x + 2)
⇒ y² – 6y – 12x – 15 = 0
which is the required eqn. of parabola.

Question 9.
The vertex at (0, 0) and the focus at (0, 1).
Solution:
Given vertex of parabola be V (0, 0) and focus be F (0, 1).

Thus the required eqn. of parabola be
x² = 4ay …(1)
since axis of parabola bey-axis, distance between focus and vertex = a = 1 – 0 = 1
∴ eqn, (1) reduces to ;
x² = 4y be the required eqn. of parabola.

Question 10.
The vertex at (0, a) and the focus at (0,0).
Solution:
Given vertex of parabola be V (0, a) and Focus F (0, 0).
Thus y-axis be the axis of downward parabola.

Hence required eqn. of parabola having vertex (0, a) be given by
(x – 0)² = – 4a (y – a)
⇒ x² = 4a(a – y)

Question 11.
The axis parallel to the x-axis, and the parabola passes through (3, 3), (6, 5) and (6, – 3).
Solution:
Let the eqn. of parabola having axis parallel to x-axis be given by
x = ay², + by + c …(1)
where a, b, c are constants,
eqn. (1) passes through the points (3, 3), (6, 5) and (6, – 3).
3 = 9a + 3b + c …(2)
6 = 25a + 5 b + c …(3)
6 = 9a – 3b + c …(4)
eqn. (2) – eqn. (4) gives ;
– 3 = 6b ⇒ b = –\(\frac { 1 }{ 2 }\)
∴ from (2); \(\frac { 9 }{ 2 }\) = 9a + c …(5)
\(\frac { 17 }{ 2 }\) = 25a + c …(6)
eqn. (6) – (5) gives;
\(\frac { 17 }{ 2 }\) – \(\frac { 9 }{ 2 }\) = 16 a
⇒ a = \(\frac { 1 }{ 4 }\)
∴ from (5); c = \(\frac { 9 }{ 2 }\) – \(\frac { 9 }{ 4 }\) = \(\frac { 9 }{ 4 }\)
∴ from (1); x = \(\frac { 1 }{ 4 }\) y² – \(\frac { 1 }{ 2 }\)y + \(\frac { 9 }{ 4 }\)
⇒ y² – 2y – 4x + 9 = 0
which is the required eqn. of Parabola.

Question 12.
The axis parallel to the y-axis and the parabola passes through the points (4, 5), (-2, 11) and (-4, 21).
Solution:
The equation of the parabola whose axis is parallel to y-axis can be taken as
y = Ax² + Bx + C …(1)
where A, B and C are arbitrary constants. Since the parabola (1) passes through the point A (4, 5).
5 = 16A + 4B + C …(2)
Clearly the points (-2, 11) and (-4, 21) lies on eqn. (1).
11 = 4A – 2B + C …(3)
21 = 16A – 4B + C …(4)
eqn. (4) – eqn. (2); we have
16 = – 8B ⇒ B = – 2
eqn. (2) – eqn. (3) gives ;
– 6 = 12A + 6B ⇒ – 6 = 12A – 12
⇒ A = \(\frac { 1 }{ 2 }\)
∴ from (2); 5 = 8 – 8 + C ⇒ C = 5
putting the values of A, B and C in eqn. (1); we have
y = \(\frac{x^2}{2}\) – 2x + 5
⇒ x² – 4x – 2y + 10 = 0
be the required eqn. of parabola.

Question 13.
The parabola y² = 4px passes through the point (3, – 2). Obtain the length of the latus rectum and the coordinates of the focus.
Solution:
Given eqn. be parabola be
y² = 4px …(1)
since eqn. (1) passes through the point (3, – 2)
∴ 4 = 4p × 3 ⇒ 12p = 4 ⇒ P = \(\frac { 1 }{ 3 }\)
∴ eqn. of parabola becomes
y² = \(\frac { 4 }{ 3 }\)x …(2)
On comparing with y² = 4ax
So eqn. (2) represents right handed parabola.
Here, 4a = \(\frac { 4 }{ 3 }\) ⇒ a = \(\frac { 1 }{ 3 }\)
∴ focus of parabola be (a, 0) i.e. \(\left(\frac{1}{3}, 0\right)\)
and length of latus-rectum = 4a = \(\frac { 4 }{ 3 }\)
∴ focus of parabola be (a, 0) i.e. \(\left(\frac{1}{3}, 0\right)\)
and length of latus-rectum = 4a = \(\frac { 4 }{ 3 }\)

Question 14.
Prove that the equation y² + 2ax + 2by + c = 0 represents a parabola whose axis is parallel to the axis of x. Find its vertex.
Solution:
Given eqn. be,
y² + 2ax + 2by + c = 0
⇒ y² + 2by + b² – b² + 2ax + c = 0
⇒ (y + b)² = – 2ax + b² – c
⇒ (y + b)² = -2a\(\left[x-\frac{b^2-c}{2 a}\right]\) …(1)
As it is of the form (y – β)² = 4a (x – a)
Thus eqn. (1) represents a parabola whose axis parallel to x-axis with vertex \(\left(\frac{b^2-c}{2 a},-b\right)\)

Question 15.
Of the parabola, 4(y – 1)² = – 7 (x – 3) find
(i) the length of the latus rectum.
(ii) the coordinates of the focus and the vertex.
Solution:
Given eqn. of parabola be
4(y – 1)² = – 7 (x – 3)
⇒ (y – 1)² = –\(\frac { 7 }{ 4 }\)(x – 3) …(1)
which is clearly represents a parabola with axes parallel to x-axis with vertex (3, 1)
putting y – 1 = Y and x – 3 = X
eqn. (1) reduces to ; Y² = –\(\frac { 7 }{ 3 }\)X,
Here 4a = \(\frac { 7 }{ 4 }\) ⇒ a = \(\frac { 7 }{ 16 }\)
∴Focus is given by (- a, 0)
i.e. X = – a and Y = 0
⇒ x – 3 = –\(\frac { 7 }{ 16 }\) and y – 1 = 0
⇒ x = \(\frac { 41 }{ 16 }\) and y = 1
i.e. focus of parabola given by (1) be \(\left(\frac{41}{16}, 1\right)\)

Question 16.
Find the vertex, focus, and directrix of the following parabolas :
(i) y² – 2y + 8x – 23 = 0
(ii) x² + 8x + 12y + 4 = 0
Solution:
(i) Given eqn. of parabola be
y² – 2y + 8x – 23 = 0
⇒ y² – 2y + 1 = – 8x + 23 + 1
⇒ (y – 1)² = – 8 (x – 3) …(1)
Transferring origin to point (3,1); we put x-3 = X; y – 1 = Y in eqn. (1); we get Y² = – 8X, which is a left handed parabola, on comparing with Y² = – 4aX i.e. 4a = 8 ⇒ a = 2
Thus, vertex is given by X = 0, Y = 0
i.e. x – 3 = 0 and y – 1 = 0
i.e. x = 3, y = 1
Thus vertex of eqn. (1) be given by (3,1). and Focus is given by X = – a, Y = 0
i.e. x – 3 = – 2 and y – 1 = 0
i.e. x = 1 and y = 1
Thus focus of eqn. (1) be given by (1, 1) and directrix is given by X = a
⇒ x – 3 = 2 ⇒ x – 5 = 0 bethe required eqn. of directrix.

(ii) Given eqn. of parabola be
x² + 8x + 12y + 4 = 0
⇒ (x2 + 8x + 16) + 12y + 4 – 16 = 0
⇒ (x + 4)² = – 12 (y – 1) …(1)
Shifting the origin to point (- 4, 1)
we put x + 4 = X; y – 1 = Y in eqn. (1).
X² = – 12Y …(2)
which represents a downward parabola.
On comparing eqn. (2) with X² = – 4aY
∴ 4a = 12 ⇒ a = 3
Thus vertex of eqn. (2) be given by
X = 0, Y = 0
i.e. x + 4 = 0 and y – 1 = 0
i.e. x = -4 and y = 1
Thus (- 4,1) be the vertex of parabola (1) Focus of eqn. (2) be given by
X = 0 and Y = – a
⇒ x + 4 = 0 and y – 1 = – 3
i.e. x = -4 and y = -2
Hence (-4, -2) be the focus of parabola (1). Thus, directrix of the parabola (2) be given by
Y = a i.e. y – 1 = 3 ⇒ y = 4
be the required directrix of parabola (1).

Question 17.
Find the vertex, focus, and directrix of the parabola (x – h)² + 4a(y – k) = 0.
Solution:
Given eqn. of parabola be
(x – h)² = – 4a(y – K) …(1)
Shifting (0, 0) to point (h, k)
putting x – h = X
and y – k = Y in eqn. (1); we have
X² = – 4aY …(2)

which represents a downward parabola vertex of eqn. (2) be given by X = 0 = Y
i.e. x – h = 0 = y – k i.e. x = h and y = k
Thus (h, k) be the vertex of parabola (1).
Focus of (2) be given by X = 0 and Y = -a
i.e. x – h – 0 and y – k = – a
i.e. x = h and y = k – a
Thus (h, k – a) be the required focus of parabola (1) and directrix of parabola (2)
be given by Y = a ⇒ y – k = a ⇒ y = k + a which is required eqn. of directrix of parabola (1).

Question 18.
Find the equation to the parabola whose axis is parallel to the y-axis and which passes through the points (0, 4), (1, 9) and (- 2, 6) and determine its latus rectum.
Solution:
Let the eqn. of parabola whose axes parallel to y-axis be given by
y = ax² + bx + c …(1)
where a, b, c are all constants
eqn. (1) passes through the points (0, 4), (1, 9) and (-2, 6).
4 = a × 0² + b × 0 + c ⇒ c = 4
9 = a + b + c ⇒ a + b = 5 ..(2)
6 = 4a – 2b + c ⇒ 4a – 2b = 2
⇒ 2a – b = 1 …(3)
On adding (2) and (3) ; we have
3a = 6 ⇒ a = 2 ∴ b = 3
putting the values a, b and c in eqn. (1); we have
y = 2x² + 3x + 4
⇒ y = 2 \(\left(x^2+\frac{3}{2} x+\frac{9}{16}\right)\) – \(\frac { 9 }{ 16 }\) + 4
⇒ y = 2 \(\left(x+\frac{3}{4}\right)^2\) + \(\frac { 55 }{ 16 }\)
⇒ \(2\left(x+\frac{3}{4}\right)^2\) = \(\left(y-\frac{55}{16}\right)\)
⇒ \(\left(x+\frac{3}{4}\right)^2\) = \(\frac { 1 }{ 2 }\)\(\left(y-\frac{55}{16}\right)\)
∴ length of latus rectum = \(\frac { 1 }{ 2 }\)

Question 19.
Find the coordinates of the point on the parabola y² = 8x whose focal distance is 8.
Solution:
Given eqn. of parabola be y² = 8x …(1)
Let (x₁, y₁) be any point on curve (1).
∴ y₁² = 8x₁ …(2)
On comparing eqn. (1) with y² = 4ax
∴ 4a = 8 ⇒ a = 2
Also, given focal distance = 8 ⇒ x_i + a = 8
⇒ x₁ = 8 – 2 = 6
∴ from (2); y₁² = 8 × 6 = 48
⇒ y₁ = ±4√3
Hence the required point on given parabola be (6, ± 4√3).

Question 20.
If the ordinate of a point on the parabola y² = 4ax is twice the latus rectum, prove that the abscissa of this point is twice the ordinate.
Solution:
Given eqn. of parabola be
y² = 4ax …(1)
Let (x₁, y₁) be any point on parabola (1).
∴ y₁² = 4ax₁ …(2)
given y₁ = 2 × length of latus rectum
= 2 × 4a = 8a
∴ from (2); (8a)² = 4ox₁
⇒ x₁ = \(\frac{64 a^2}{4 a}\) = 16a ⇒ x₁ = 2 (8a) = 2y₁
Thus abscissa of this point is twice the ordinate.

Question 21.
Find the equation of the parabola whose focus is at the origin, and whose directrix is the line y – x = 4. Find also the length of the latus rectum, the equation of the axis, and the coordinates of the vertex.
Solution:
Given (0, 0) be the focus of required parabola whose directrix is the line
y – x – 4 = 0
Let P (x, y) be any point on parabola.
Then by def. | PF | = | PM |

On squaring both sides ; we have
2 [x² + y²] = (y – x – 4)²
⇒ 2(x² + y²) = y² + x² – 2xy + 8x – 8y + 16
⇒ x² + y² + 2xy – 8x + 8y – 16 = 0
which is the required eqn. of parabola, length of latus rectum = 4a = 2 × 2a
= 2 (length of ⊥ drawm from focus (0, 0) to directrix y – x – 4 = 0)

Clearly axis be the line through the focus F (0, 0) and ⊥ to directrix
y – x – 4 = 0 …(1)
eqn. of axis be given by
x + y + k = 0
it pass through F (0, 0). ∴ k = 0
Thus, x + y = 0 …(2)
be the eqn. of axis of parabola.
Let Z be the point of intersection of directrix and axis of parabola. So we solve eqn. (1) and eqn. (2) simultaneously ; we have
y = 2 and x = – 2
∴ Coordinates of point Z are (- 2, 2). and vertex (α, β) be the mid-point of FZ.
α = \(\frac{-2+0}{2}\) and β = \(\frac{2+0}{2}\)
i.e. a = – 1 and β = 1
Thus, required vertex of parabola be (-1, 1).

Question 22.
The directrix of a conic section is the straight line 3x – 4y + 5 = 0 and the focus is (2, 3). If the eccentricity e is 1, find the equation to the conic section. Is the conic section a parabola ?
Solution:
Since eccentricity of conic section be 1.
∴ conic section is a parabola,
given eqn. of directrix be 3x – 4y + 5 = 0 and focus be F (2, 3). Let P (x, y) be any point on parabola.
Then by def. | PF | = | PM |

On squaring both sides ; we have
25 [(x – 2)² + (y – 3)²] = (3x – 4y + 5)²
⇒ 25 [x² + y² – 4x – 6y + 13]
= 9x² + 16y² + 25 – 24xy – 40y + 30x
⇒ 16x² + 9y² + 24xy – 130x – 110y + 300 = 0
which is the required eqn. of parabola.

Question 23.
Find the equation to the parabola whose focus is (-2, 1) and directrix is 6x – 3y = 8.
Solution:
Given focus be (- 2, 1) and eqn. of directrix be 6x – 3y – 8 = 0
Let P (x, y) be any point on parabola.
Then by def. | PF | = | PM |

On squaring both sides ; we have
45 [(x + 2)² + (y – 1)²] = (6x -3y- 8)²
⇒ 45 [x² + 4x + y² – 2y + 5]
= 36x² + 9y² + 64 – 36xy + 48y – 96x
⇒ 9x² + 36y² + 36xy + 276x – 138y + 161 = 0
which is the required eqn. of parabola.

Question 24.
The length of the latus rectum of the parabola whose focus is (3, 3) and directrix is 3x – 4y – 2 = 0 is
(a) 2
(b) 1
(c) 4
(d) None of these
Solution:
Required length of latus rectum = 4a
= 2 × (length of 1 draw from focus (3, 3) to directrix 3x – 4y – 2 = 0)